Now on to the fun part! Let’s take a look at some of the fundamental equations and break them down. Our goal is to understand the rationale behind the equations, increasing our confidence in applying them to save Dr. Vanko. But keep in mind, a miscalculation could lead to grave outcomes. (no pressure )
Loading Dose (LD)
20-25 mg/kg x Actual body weight
Volume of Distribution (Vd)
0.7 L/kg x Actual body weight
Elimination Constant (Ke)
0.00083 (CrCl) + 0.0044
|In(desired peak/desired trough)/Ke + t’|
Maintenance dose (MD)
LD x 1-e-KeT
(Dose/t’)(1-e-Ke(t’)) / Vd(Ke)(1-e-Ke(T))
Vd = volume of distribution, Ke = elimination constant, T= interval, t’ = infusion time
LD = 20-25 mg/kg OR LD = desired peak x Vd
Loading doses are used to get the patient to therapeutic drug concentrations quicker. The thought behind the practice is to ‘fill the tank‘ since we are starting from nothing. Loading doses are recommended in severe infections (e.g. sepsis, bacteremia, meningitis) where rapid attainment of therapeutic drug concentrations are desired to prevent adverse outcomes.
In general, we want to use the lower loading dose range (20 mg/kg dose) in indications with a lower trough goal (10-15 mcg/mL) and the higher range (25 mg/kg) in indications with a higher trough goal (15-20 mcg/mL).
LD = 20 mg/kg x total body weight*
LD = 25 mg/kg x total body weight*
*Total body weight = actual body weight
Another method of calculating the loading dose is with the equation LD = desired peak x Vd. Taking a closer look at this equation, you will find that it is familiar. It is the same equation as the volume of distribution equation we previously reviewed, just rearranged to find the dose.
or Dose (mg) = Concentration (mg/mL) x Volume of distribution
In this example, knowing the population estimated volume of distribution and the desired concentration (or peak), you are trying to find the dose (or amount of sugar dissolved in the container as discussed in lesson #1) to achieve that.
This equation allows you to target a goal peak. Common goal peaks range from 30-40 mcg/mL.
It is important to follow your institution’s policy as each institution will have specific indications for loading as well as the maximum dose allowed.
Volume of Distribution
Vd = 0.7 L/kg
As discussed previously, the equation for determining the volume of distribution is:
Why then is the equation for the volume of distribution of vancomycin 0.7 L/kg?
This equation was derived from population kinetic studies, in which a large sample of patients was administered vancomycin and then plasma concentrations were drawn and compared to the doses given. The average volume of distribution per weight of the patients in this study resulted in the commonly used Vd equation 0.7 L/kg.
The volume of distribution can vary significantly between patients with ranges reported in the literature from 0.5-1 L/kg, making determinations of an exact dose rather tricky. That is why population kinetics is an estimation all around! 😎
Population Ke = 0.00083 (CrCl) + 0.0044
Ke is the elimination constant or the fraction of vancomycin that is eliminated from the body per unit of time. Because vancomycin is primarily eliminated via the kidneys, the elimination constant is directly related to Creatinine Clearance.
This equation is derived similarly to the volume of distribution equation. Population kinetic studies were conducted in a random group of people with Ke plotted against Creatinine Clearance resulting in a line of best fit (y = mx + b) shown below.
Estimated half-life (t1/2) = 0.693/Ke
Half-life is defined as the time it takes for the concentration of the drug to decrease by 50% in the body as shown in the graph below. After 4-5 half-lives, drug concentrations reach negligible levels in the body.
If you have taken chemistry, you may remember learning how to calculate the half-life of a radioactive substance (ex: uranium) as it decays or using half-life to determine the age of a piece of coal. The half-life equation for vancomycin is essentially the same.
The numerator is the natural log of 2. Knowing what the patient’s elimination constant is, you are calculating how long it will take the serum concentration of drug to decrease by half. A simplified way to think of it is shown below.
Goal peak = 30 mcg/mL —> after 1 half-life concentrations will decrease by 50% —> 15 mcg/mL
Redose the patient after each half-life to keep drug concentrations within the goal trough of 15 mcg/mL
The half-life can be used to determine the dosing interval.
General rule: you do not want to use an interval that is LESS than your calculated half-life as this can lead to overaccumulation and high drug concentrations (example: using a dosing interval of every 12 hours when your t1/2 was 16 hours). However, if you make the clinical judgment to use an interval that is less than your calculated half-life, decrease your maintenance dose to 10-12 mg/kg to ensure trough levels stay within the goal. If you’re unsure, you can input your calculated maintenance dose and interval into the estimated peak and trough equations to get a better idea.
Estimated interval = In (desired peak/desired trough)/ (Ke + t’)
Where t’ = infusion time is usually 1 hour
To calculate the interval, the equation requires you to input the desired peak (Cpeak) of 30-40 mcg/L.
Since the half-life and interval equations are both dependent on clearance (Ke) to determine how much of the drug will be eliminated from the body, they produce similar results. For the most part, if you memorize the half-life equation (t1/2 = 0.693/Ke), it should be around the same as your interval equation. You just have to add in your infusion time. Let’s test it out for fun!
It is important to round your calculations to their nearest standard intervals (e.g. every 8 hours, 12 hours, or 24 hours).
MD = LD x 1-e-KeT
The maintenance dose equation calculates the amount of drug lost from the loading dose at the end of the interval previously determined. This amount becomes your maintenance dose.
- You administer a loading dose of 2,000 mg.
- Based on your calculations, the patient’s elimination constant and interval come out to be Ke = 0.073 and T = 12 hours. Remember that the elimination constant and interval equations have already taken into account your desired peak (30 mcg/mL) and trough (15 mcg/mL).
- Approximately 12 hours after you have given your 2,000 mg loading dose, the amount of drug eliminated will be 1,167 mg, where the drug concentration is expected to be around 15 mcg/mL.
- Your maintenance dose would be 1,250 mg rounded to the nearest 250 mg increments.
Peak (Cpeak) = Dose/t’)(1-e-Ke(t’)) / Vd(Ke)(1-e-Ke(T))
After calculating your maintenance dose and interval rounded to the nearest increments, you want to determine what your Cpeak will be. This requires you to utilize the Cpeak equation above. It’s a pretty busy equation so it’s okay if you’re eyes are spinning out of your head now.
Simply put, the equation is a component of the time it takes for the drug to infuse into the blood circulation over the elimination of the drug from the body. When these two components reach equilibrium you have to achieve maximum blood concentration in the body. The amount goes in = the amount eliminated 🧘.
Trough (Ctrough) = Cpeak (e-Ke(T-t’))
After calculating the estimated peak (Cpeak), you can calculate what the trough level will be at the end of your dosing interval.
This gives you the estimated trough at steady state.
Great job finishing Lesson #3: Vancomycin Equations! You are over halfway done with your mission. Now on to the next quiz to see how well you know these equations. If you obtain 80% on the quiz, you will unlock a special token (vancomycin equation cheat sheet) to help you with the rest of your journey. Good luck!